Let `x_(1),x_(2),x_(3),....` be terms of an AP, if `(x_(1)+x_(2)+...+x_(n))/(x_(1)+x_(2)+...+x_(m))=(n^(2))/(m^(2)).(n!=m)," then "(x_(8))/(x_(23))=?`

If x_(1),x_(2),x_(3)...x_(n) are in H.P,then prove that x_(1)x_(2)+x_(2)x_(3)+xx x-3x_(4)+...+x_(n-1)x_(n)=(n-1)x_(1)x_(n)

(m)/(n)x^(2)+(n)/(m)=1-2x

If x_(1), x_(2), x_(3) .................x_(n) are in H.P.then x_(1)x_(2)+x_(2)x_(3)+.......+x_(n-1)x_(n) =

If x_(1),x_(2),x_(3),"……" are in A.P., then the value of 1/(x_(1)x_(2))+1/(x_(2)x_(3))+1/(x_(3)x_(4))+"……"1/(x_(n-1)x_(n)) is :

If x_(1),x_(2),x_(3),......x2_(n) are in A.P, then sum_(r=1)^(2n)(-1)^(r+1)x_(r)^(2) is equal to

Each of the numbers x_(1),x_(2),x_(3),...x_(n),n>=4 is equal to 1 or -1. Suppose x_(1)x_(2)x_(3)x_(4)+x_(2)x_(3)x_(4)x_(5)+x_(3)x_(4)x_(5)x_(6)+....x_(n-3)x_(n-2)x_(n-1)x_(n)+x_(n-2)x_(n-1)x_(n)x_(1)+x_(n-1)x then